(a) If $x = e^{x/y}$, prove that $\dfrac{dy}{dx} = \dfrac{\log x - 1}{(\log x)^2}$. OR (b) Check the differentiability of $f(x) = \begin{cases} x^2 + 1, & 0 \le x < 1 \\ 3 - x, & 1 \le x \le 2 \end{cases}$ at $x = 1$.

(a) If x = ex/y, prove that (dy)/(dx) = (log x - 1)/((l — CBSE Class 12 Mathematics 2024 PYQ

Question

x = e^{x/y}

(OR)

(b) Check the differentiability of

f(x) = \begin{cases} x^2 + 1, & 0 \le x < 1 \\ 3 - x, & 1 \le x \le 2 \end{cases}

x = 1

CBSECBSE Class XII Board 2024Subjective· 2mImportance★★★★★

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Part (a)Concept understanding — Implicit Differentiation

Implicit Differentiation

When $y$ isn't alone

You can differentiate $y=x^2+3x$ term by term because $y$ is written explicitly in terms of $x$ . But an equation like $x^2+y^2=25$ , or $x^3+y^3=6xy$ , does not give $y$ by itself — solving for $y$ is messy or downright impossible.

Implicit differentiation finds $\dfrac{dy}{dx}$ without isolating $y$ : treat $y$ as an unknown function of $x$ , differentiate the whole equation as it stands, then solve for $\dfrac{dy}{dx}$ .

The one key move: $y$ is really $y(x)$

Wherever $y$ appears, picture $y(x)$ hiding inside. Differentiating a $y$ -term therefore needs the chain rule, which tacks on a factor of $\dfrac{dy}{dx}$ :

\frac{d}{dx}\big(y^2\big)=2y\,\frac{dy}{dx}.

That extra $\dfrac{dy}{dx}$ on every $y$ -term is the whole trick.

The procedure

x

Worked example

For $x^2+y^2=25$ :

2x+2y\frac{dy}{dx}=0 \quad\Rightarrow\quad \frac{dy}{dx}=-\frac{x}{y}.

The answer naturally contains both $x$ and $y$ — that is normal here. To get the slope at a point on the curve, substitute the coordinates after differentiating; there is no need to solve for $y$ first. …

Part (b)Concept understanding — Continuity Differentiability Relationship

How Continuity and Differentiability Are Related

Two properties describe how "well-behaved" a function is at a point. Continuity means the graph has no break there — you can draw through the point without lifting your pen. Differentiability means the graph is smooth there — it has one definite tangent line, so a well-defined slope $f'(a)$ . This concept is about the exact link between the two.

Important

The theorem: If $f$ is differentiable at $x = a$ , then $f$ is continuous at $x = a$ .

Why differentiability forces continuity

If $f'(a)$ exists, then

\lim_{x\to a}\big(f(x)-f(a)\big)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}\cdot (x-a)=f'(a)\cdot 0=0.

So $\lim_{x\to a}f(x)=f(a)$ , which is exactly continuity at $a$ . A curve that has a tangent cannot also have a jump — a break would send the difference quotient to infinity and the derivative would not exist.

The converse is FALSE

Continuity does not guarantee differentiability. A graph can be unbroken yet still have a sharp corner, and a corner has no single tangent.

Watch out

The classic counterexample is $f(x)=|x|$ at $x=0$ . It is continuous there ( $\lim_{x\to0}|x|=0=f(0)$ ), but the slope from the left is $-1$ and from the right is $+1$ . Since these disagree, $f'(0)$ does not exist.

Putting it together

Differentiable at $a$ $\Rightarrow$ continuous at $a$ .
Continuous at $a$ $\not\Rightarrow$ differentiable at $a$ .
Not continuous at $a$ $\Rightarrow$ not differentiable at $a$ (the contrapositive of the theorem). …

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Part (a): taking logs turns $x=e^{x/y}$ into $y=\dfrac{x}{\log x}$ , and the quotient rule gives $\dfrac{dy}{dx}=\dfrac{\log x-1}{(\log x)^2}$ . Part (b): $f$ is continuous at $x=1$ but LHD $=2\ne$ RHD $=-1$ , so it is not differentiable there.

Part (a)

The unknown $y$ sits inside an exponent, so we free it with logarithms.

1. Take natural logs of $x=e^{x/y}$ :

\log x=\frac{x}{y}.

2. Solve for $y$ explicitly: multiply by $y$ and divide by $\log x$ :

y\log x=x\ \Rightarrow\ y=\frac{x}{\log x}.

3. Differentiate by the quotient rule with $u=x$ , $v=\log x$ ( $u'=1$ , $v'=\tfrac1x$ ):

$\frac{dy}{dx}=\frac{u'v-uv'}{v^2}=\frac{(1)(\log x)-x\cdot\frac1x}{(\log x)^2}=\frac{\log x-1}{(\log x)^2}.$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 33 on this concept.

CBSE 2026Set 65/2/11 markMCQ
$e^{-x} + e^{-y} = 2$
›Reveal solutionSolution
To find $\frac{dy}{dx}$ for an implicitly defined function, we differentiate both sides of the equation with respect to $x$ , treating $y$ as a function of $x$ and applying the chain rule. The result is $\boxed{-e^{y-x}}$ .

When an equation relates $x$ and $y$ but does not explicitly express $y$ as a function of $x$ (like $y = f(x)$ ), we use a technique called implicit differentiation to find $\frac{dy}{dx}$ . The core idea is that even though $y$ isn't isolated, it is still a function of $x$ .

This means that when we differentiate a term involving $y$ with respect to $x$ , we must apply the chain rule. For example, if we differentiate $g(y)$ with respect to $x$ , we get $\frac{d}{dx}[g(y)] = g'(y) \cdot \frac{dy}{dx}$ . This $\frac{dy}{dx}$ term is crucial and often the source of errors if overlooked.

Let's apply this to the given equation.

Differentiate both sides of the equation with respect to $x$ . The given equation is $e^{-x} + e^{-y} = 2$ . We apply the derivative operator $\frac{d}{dx}$ to every term:

$\frac{d}{dx}(e^{-x}) + \frac{d}{dx}(e^{-y}) = \frac{d}{dx}(2)$

Evaluate each derivative.

For the first term, $\frac{d}{dx}(e^{-x})$ :

Using the chain rule, if $u = -x$ , then $\frac{du}{dx} = -1$ .

So, $\frac{d}{dx}(e^{-x}) = e^{-x} \cdot \frac{d}{dx}(-x) = e^{-x} \cdot (-1) = -e^{-x}$ .

For the second term, $\frac{d}{dx}(e^{-y})$ :

This is where implicit differentiation comes in. We treat $y$ as a function of $x$ .

Using the chain rule, if $v = -y$ , then $\frac{dv}{dx} = \frac{d}{dx}(-y) = -1 \cdot \frac{dy}{dx}$ .

So, $\frac{d}{dx}(e^{-y}) = e^{-y} \cdot \frac{d}{dx}(-y) = e^{-y} \cdot \left(-\frac{dy}{dx}\right) = -e^{-y} \frac{dy}{dx}$ .

Watch out

A common mistake is to forget the $\frac{dy}{dx}$ term when differentiating expressions involving $y$ with respect to $x$ . Remember, $y$ is a function of $x$ .

For the right-hand side, $\frac{d}{dx}(2)$ : …
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CBSE 2026Set A1 markMCQ
$y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \ldots}}}$
(a) $\frac{1}{2y-1}$
(b) $\frac{\cos x}{2y-1}$
(c) $\frac{\sin x}{2y-1}$
(d) $\frac{2y-1}{\cos x}$
›Reveal solutionSolution

$\frac{dy}{dx} = \frac{\cos x}{2y-1}$ .

The infinite nested radical satisfies $y = \sqrt{\sin x + y}$ , so

$y^2 = \sin x + y.$

Differentiate both sides implicitly with respect to $x$ :

$2y\frac{dy}{dx} = \cos x + \frac{dy}{dx}.$

Collect $\frac{dy}{dx}$ : …

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CBSE 2026Set A1 markMCQ
$x^n + y^n = a^n$
(a) $-\frac{x^{n-1}}{y^{n-1}}$
(b) $\frac{x^{n-1}}{y^{n-1}}$
(c) $-\frac{y^{n-1}}{x^{n-1}}$
(d) $nx^{n-1}$
›Reveal solutionSolution

$\frac{dy}{dx} = -\frac{x^{n-1}}{y^{n-1}}$ .

Differentiate $x^n + y^n = a^n$ implicitly ( $a$ constant):

$n x^{n-1} + n y^{n-1}\frac{dy}{dx} = 0.$

Solve: …

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CBSE 2026Set ANNUAL1 markMCQ
$2x + 3y = \sin y$
(a) $\frac{3}{\sin y - 2}$
(b) $\frac{2}{\cos y - 3}$
(c) $\frac{\cos y + 3}{2}$
(d) $\frac{2}{\cos y}$
›Reveal solutionSolution

Differentiate both sides with respect to $x$ , treating $y$ as a function of $x$ , then solve for $dy/dx$ .

$2x+3y=\sin y$

Differentiating: $2 + 3\dfrac{dy}{dx} = \cos y\,\dfrac{dy}{dx}$

$2 = \dfrac{dy}{dx}(\cos y - 3)$

…

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CBSE 2026Set ANNUAL1 mark
$\dfrac{dy}{dx}$
›Reveal solutionSolution

Differentiate both sides of $2x+3y=\sin y$ with respect to $x$ (using the chain rule for the $y$ -terms), then collect $\dfrac{dy}{dx}$ on one side.

Given: $2x + 3y = \sin y$

Differentiate both sides w.r.t. $x$ :

$\frac{d}{dx}(2x) + \frac{d}{dx}(3y) = \frac{d}{dx}(\sin y)$

$2 + 3\frac{dy}{dx} = \cos y \cdot \frac{dy}{dx}$

Collect all $\dfrac{dy}{dx}$ terms on one side: …

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CBSE 2026Set ANNUAL1 mark
$Q. Write True or False: Every differentiable function is continuous, but the converse is not true.$
›Reveal solutionSolution

True - differentiable $\Rightarrow$ continuous, converse false.

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Assertion (A): Every differentiable function is continuous. Reason (R): Every continuous function is differentiable. Choose the correct option:$
(a) Both A and R are correct and R is the correct explanation of A.
(b) Both A and R are correct but R is not the correct explanation of A.
(c) A is correct but R is incorrect.
(d) Both A and R are incorrect.
›Reveal solutionSolution

Differentiability implies continuity, but the converse is false.

A is TRUE: Every differentiable function is continuous — this is a standard theorem.

…

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CBSE 2025Set ANNUAL1 markMCQ
$x^2+y^2=2$
(a) $\dfrac{1-2x}{2y}$
(b) $\dfrac{2y}{1-2x}$
(c) $-\dfrac{x}{y}$
(d) $-\dfrac{y}{x}$
›Reveal solutionSolution

Differentiate $x^2+y^2=2$ implicitly with respect to $x$ , treating $y$ as a function of $x$ .

$\dfrac{d}{dx}(x^2+y^2)=\dfrac{d}{dx}(2)$

$2x+2y\dfrac{dy}{dx}=0$ …

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CBSE 2025Set ANNUAL1 mark
$\dfrac{dy}{dx}$
›Reveal solutionSolution

Differentiate both sides with respect to $x$ , treating $y$ as a function of $x$ .

Start from $ax+by^2=\cos y$ and differentiate w.r.t. $x$ :

$\frac{d}{dx}(ax)+\frac{d}{dx}(by^2)=\frac{d}{dx}(\cos y)$

$a+2by\frac{dy}{dx}=-\sin y\frac{dy}{dx}.$

Gather the $\dfrac{dy}{dx}$ terms:

$2by\frac{dy}{dx}+\sin y\frac{dy}{dx}=-a$

$\frac{dy}{dx}\big(2by+\sin y\big)=-a.$

Hence

$\frac{dy}{dx}=\frac{-a}{2by+\sin y}.$

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The value of dy/dx at (4, 1) of y³ - \sqrtx = 5 is ......................$
(a) 5/4
(b) 1/12
(c) 1/24
(d) 1/3
›Reveal solutionSolution

Differentiate the implicit relation $y^3 - \sqrt{x} = 5$ term by term with respect to $x$ , then substitute the point $(4, 1)$ .

Given: $y^3 - \sqrt{x} = 5$

Step 1 — differentiate implicitly w.r.t. $x$ :

$3y^2 \frac{dy}{dx} - \frac{1}{2\sqrt{x}} = 0$

Step 2 — solve for $dy/dx$ :

$\frac{dy}{dx} = \frac{1}{2\sqrt{x}\cdot 3y^2} = \frac{1}{6y^2\sqrt{x}}$

…

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CBSE 2024Set D1 markMCQ
$y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \ldots \text{ to } \infty}}}$
(a) $\frac{\sin x}{2y - 1}$
(b) $\frac{\cos x}{y - 1}$
(c) $\frac{\cos x}{2y - 1}$
(d) $\frac{1}{2y - 1}$
›Reveal solutionSolution

$\frac{dy}{dx} = \frac{\cos x}{2y - 1}$ .

The infinite nested radical satisfies $y = \sqrt{\sin x + y}$ because the expression inside the outer root repeats. Square both sides:

$y^2 = \sin x + y.$

Differentiate implicitly with respect to $x$ :

$2y\frac{dy}{dx} = \cos x + \frac{dy}{dx}.$ …

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CBSE 2024Set ANNUAL1 markMCQ
$2x + 8y = \sin x$
(a) $\dfrac{\sin x - 2}{8}$
(b) $\dfrac{\cos x - 2}{8}$
(c) $\dfrac{\cos x + 2}{2}$
(d) $\dfrac{\cos x + 2}{3}$
›Reveal solutionSolution

Differentiate both sides of $2x+8y=\sin x$ implicitly with respect to $x$ and isolate $\dfrac{dy}{dx}$ .

$2x + 8y = \sin x$

Differentiating both sides w.r.t. $x$ :

$2 + 8\dfrac{dy}{dx} = \cos x$

…

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Q.(a) If x=ex/yx = e^{x/y}x=ex/y, prove that dydx=log⁡x−1(log⁡x)2\dfrac{dy}{dx} = \dfrac{\log x - 1}{(\log x)^2}dxdy​=(logx)2logx−1​.

Part (a)

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Q.(a) If $x = e^{x/y}$ , prove that $\dfrac{dy}{dx} = \dfrac{\log x - 1}{(\log x)^2}$ .

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